A metal rod 0.60 m long turns about one end, in a plane at right angles to a uniform field of 0.40…

Step 1. The angular speed. $\omega = 2\pi f = 2\pi \times 5.0\ \text{s}^{-1} = 31.4\ \text{rad/s}$. The speed of the free end is $\omega l = 31.4\ \text{rad/s} \times 0.60\ \text{m} = 18.8\ \text{m/s}$.

Step 2. The emf. $\varepsilon = \tfrac{1}{2}B\omega l^2 = \tfrac{1}{2} \times 0.40\ \text{T} \times 31.4\ \text{rad/s} \times (0.60\ \text{m})^2 = 2.26\ \text{V}$.

Step 3. The middle point. The part from the pivot to the middle has length $0.30$ m. So $\tfrac{1}{2}B\omega (0.30\ \text{m})^2 = 0.565\ \text{V}$.